Major remarks.

Quote:

Hence, $\displaystyle a$ is an upper bound for $\displaystyle \{f(x)|x<a\}$.

Should be f(a), not a.

The second claim is not really proved.

Quote:

Then, since $\displaystyle x<a$ for all $\displaystyle x,a\in I$, then $\displaystyle a$ must be the least upper bound for $\displaystyle \{f(x)|x<a\}$.

Even the claim that f(a) (rather than a) is the least upper bound for $\displaystyle \{f(x)\mid x<a\}$ is incorrect. Consider

$\displaystyle f(x)=

\begin{cases}

x & x<a\\

x+1 & x\ge a

\end{cases}

$

Then $\displaystyle \sup\{f(x)\mid x<a\}=a$, but $\displaystyle f(a)=a+1$.

Even if $\displaystyle f(a)=\sup\{f(x)\mid x<a\}$, how does it help prove the following? Quote:

Hence, $\displaystyle lim_{x\nearrow a}f(x)=sup\{f(x)|x<a\}$.

After all, $\displaystyle \lim_{x\nearrow a}f(x)=f(a)$ only if f is left-continuous at a. In my opinion, there is at least not enough details here that are expected from a proof of such a simple fact.

Minor remarks.

Quote:

Then, let $\displaystyle x<a$ for $\displaystyle a\in I$ and all $\displaystyle x\in I$.

You can't expect that x < a for all x in I. Since a is already fixed in the problem statement, this phrase should say, "Let x < a for *some* x in I".

Quote:

Thus, since $\displaystyle x<a\Rightarrow f(x)<f(a)$ for all $\displaystyle a,x\in I$.

Again, it is not the case that f(x) < f(a) holds for *all* a, x in I.